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May 29, 2021 at 0:01 comment added Eric Towers Certainly, there are denser ways to pack circles. What you are lacking is a proof that the denser packings also don't fit 101 circles in the big circle.
May 28, 2021 at 23:24 history edited RobPratt
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May 27, 2021 at 7:55 comment added Jaap Scherphuis To prove the statement using the pigeonhole principle, you want to split the disc into 100 (or fewer) regions. That way you know that with 101 points, there must be some region with at least two points in it. For this to work as a proof of the statement, you need those regions to have diameter less than 2, so that any region with 2 (or more) points forces two points to be less than 2 apart. As Mindlack points out, your 81 squares have diameter of 2.82..., so do nothing to prove the statement.
May 27, 2021 at 7:35 comment added KlesierTheSurvivor well each dot on the 2x2 square is exactly a distance apart of 2, and so the diagonal will always be more than 2.
May 27, 2021 at 7:29 comment added Aphelli This isn’t very clear, unfortunately... but in your tiling of the $18 \times 18$ square with $81$ $2 \times 2$ squares, please note that two elements in the same square need not be at distance less than two (consider the opposite corners).
May 27, 2021 at 7:08 review First posts
May 27, 2021 at 7:08
May 27, 2021 at 7:04 history asked KlesierTheSurvivor CC BY-SA 4.0